Differential Equations
Differential Equations with Initial Conditions
Grade 12
Question:
<p>Let \(f : [0,1] \to \mathbb{R}\) be such that \(f(xy) = f(x)\cdot f(y)\), for all \(x, y \in [0,1]\), and \(f(0) \neq 0\). If \(y = y(x)\) satisfies the differential equation, \(\dfrac{dy}{dx} = f(x)\) with \(y(0) = 1\), then \(y\!\left(\dfrac{1}{4}\right) + y\!\left(\dfrac{3}{4}\right)\) is equal to _____.</p>
Step-by-Step Solution
Key Concept: From the functional equation f(xy) = f(x)·f(y) with f(0) ≠ 0, setting y=0 gives f(0) = f(x)·f(0), so f(x) = 1 for all x. Thus the differential equation becomes dy/dx = 1, which is trivial to solve.
<p><strong>Step 1:</strong> Analyze the functional equation f(xy) = f(x)·f(y) for x, y ∈ [0,1].</p><p><strong>Step 2:</strong> Set y = 0: f(x·0) = f(x)·f(0) ⟹ f(0) = f(x)·f(0).</p><p><strong>Step 3:</strong> Since f(0) ≠ 0, we can divide by f(0): 1 = f(x). Thus f(x) = 1 for all x ∈ [0,1].</p><p><strong>Step 4:</strong> The differential equation becomes dy/dx = 1 with y(0) = 1.</p><p><strong>Step 5:</strong> Integrating: y(x) = x + C. Using y(0) = 1: C = 1, so y(x) = x + 1.</p><p><strong>Step 6:</strong> Calculate y(1/4) + y(3/4) = (1/4 + 1) + (3/4 + 1) = 5/4 + 7/4 = 12/4 = 3.</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: 3